Let $X,Y$ are independent uniform random variables on $[0,1]$.

How do I find firstly the joint density function then secondly the joint distribution function.

I know $$f_X(x)=\begin{cases}1&:&x\in[0,1]\\0 &:&\text{otherwise}\end{cases}$$

$$f_Y(y)=\begin{cases}1&:&y\in[0,1]\\0 &:&\text{otherwise}\end{cases}$$

but I am unsure how to get the joint density function firstly then I am not 100% sure how to get the distribution (although I suspect you integrate the density function).


By independence, the joint density function is $f_{X,Y}(x,y) = f_X(x) f_Y(y)$. Be sure to incorporate the information about the support of $f_X$ and $f_Y$ (where it is zero and where it is nonzero).

Similarly, the joint distribution function is $P(X\le x, Y \le y) = P(X \le x) P(Y \le y)$ by independence. Again, be careful: $P(X\le x)$ is piecewise linear.

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  • $\begingroup$ Is it $$f_{X,Y}(x,y)=\begin{cases}1&:&(x,y)\in[0,1]^2\\0 &:&\text{otherwise}\end{cases}$$. And $$F_{X,Y}(x,y)=\begin{cases}1&:&x,y \geq 1 \\0 &:&~\text{if} ~ x ~\text{or}~ y \leq 0 \\ xy &:&~\text{if}~ (x,y) \in [0,1]^2\end{cases}$$? $\endgroup$ – Ryan S Oct 25 '16 at 22:31
  • $\begingroup$ @RyanS You're missing some cases for $F_{X,Y}$, like if $x \ge 1$ and $y \in [0,1]$. $\endgroup$ – angryavian Oct 26 '16 at 0:41
  • $\begingroup$ @RyanS Specifically these blanks: $$F_{X,Y}(x,y)=\begin{cases} 0 &:& x\leqslant 0 ~\vee~ y\leqslant 0 \\ xy &:& (x,y)\in(0;1)^2 \\ \boxed{\quad?} & : & x\in(0;1)~\wedge~ 1\leqslant y \\ \boxed{\quad?} &:& 1\leqslant x ~\wedge~ y\in(0;1) \\ 1 &:& (x,y)\in[1;\infty)^2 \end{cases}$$ $\endgroup$ – Graham Kemp Oct 26 '16 at 2:56
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    $\begingroup$ @GrahamKemp Would it just be $x$ and then $y$ in the blanks then? Thank you. $\endgroup$ – Ryan S Oct 26 '16 at 8:41
  • $\begingroup$ Indeed; that it is. @RyanS $\endgroup$ – Graham Kemp Oct 26 '16 at 9:05

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